In class we proved that $x \mapsto \int_{x_0}^x g(t) dt$ is absolutely continuous for $g \in L^1(I)$, where $I$ is any open interval (but not necessarily bounded).
For the proof of a theorem, the professor used that for $u \in W^{1, p}(I)$ with $1 \leq p \leq + \infty$, the function $x \mapsto \int_{x_0}^x u'(t)dt$ is absolutely continuous. Here $u'(t)$ denotes the weak derivative.
What confused me about that step, is that we didn't assume that the interval $I$ is bounded. If we did so, we could embed $W^{1, p}(I)$ into $W^{1, 1}(I)$ using the Hölder-inequality.
However, for $I = \mathbf{R}$ couldn't it be that $u' \notin L^1(\mathbf{R})$ and then that the function $x \mapsto \int_{x_0}^x u'(t) dt$ is not absolutely continuous?
Or can I embed $W^{1, p}(\mathbf{R})$ into $W^{1, 1}(\mathbf{R})$ anyway?
The Sobolev space $W^{1,q}$, $q>1$, does not embed in $W^{1,1}$ on unbounded domains. For example, $f(x) = \sin x /x$ does not belong to $W^{1,1}(\mathbb{R})$ although it belongs to $W^{1,q}(\mathbb{R})$ for all $q>1$.
But for the proof of absolute continuity this does not matter. Let $v(x) = \int_{x_0}^x u'(t)\,dt$. The definition involves a finite sequence of disjoint intervals $(x_k,y_k)$ such that $\sum_k(y_k-x_k)\le \delta$. By Hölder's inequality on the set $E = \bigcup_k (x_k,y_k)$,
$$ \int_E |u'| \le \left(\int_E 1 \right)^{1/p} \left(\int_E |u'|^q\right)^{1/q} \le \delta^{1/p} \left(\int_E |u'|^q\right)^{1/q} $$ where $p$ is the conjugate exponent for $q$.
Hence $\sum_k |u(x_k)-u(y_k)|$ is small for two reasons:
Terminology
Absolute continuity of a function is often understood as a local property; i.e., a function $f\colon \mathbb{R}\to\mathbb{R}$ is called absolutely continuous if its restriction to every bounded interval is absolutely continuous. If this is the approach taken in your course, the point is moot. Otherwise, see the proof above.